Find the slope and y-intercept of the line that is ${\text{perpendicular}}$ to $\enspace {y = x + 2}\enspace$ and passes through the point ${(5, -7)}$. ${1}$ ${2}$ ${3}$ ${4}$ ${5}$ ${6}$ ${7}$ ${8}$ ${9}$ ${\llap{-}2}$ ${\llap{-}3}$ ${\llap{-}4}$ ${\llap{-}5}$ ${\llap{-}6}$ ${\llap{-}7}$ ${\llap{-}8}$ ${\llap{-}9}$ ${1}$ ${2}$ ${3}$ ${4}$ ${5}$ ${6}$ ${7}$ ${8}$ ${9}$ ${\llap{-}2}$ ${\llap{-}3}$ ${\llap{-}4}$ ${\llap{-}5}$ ${\llap{-}6}$ ${\llap{-}7}$ ${\llap{-}8}$ ${\llap{-}9}$
Answer: Lines are considered perpendicular if their slopes are negative reciprocals of each other. The slope of the blue line is ${1}$ , and its negative reciprocal is ${-1}$ Thus, the equation of our perpendicular line will be of the form $\enspace {y = -x + b}\enspace$ We can plug our point, $(5, -7)$ , into this equation to solve for ${b}$ , the y-intercept. $-7 = {-}(5) + {b}$ $-7 = -5 + {b}$ $-7 + 5 = {b} = -2$ The equation of the perpendicular line is $\enspace {y = -x - 2}\enspace$. ${m = -1, \enspace b = -2}$